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Centripetal Acceleration & Resultant Force

Key idea: Explain inward acceleration in uniform circular motion and apply a = v²/r = rω² and the corresponding resultant-force equations.

  • GCE A-Level H2 Physics 2027
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Learning objectives

  • Explain and apply centripetal acceleration and resultant-force relationships.

1. Definitions (Must Know)

A. Uniform circular motion

Uniform circular motion is motion in a circle at constant speed (constant magnitude of velocity).

B. Centripetal acceleration, a_c

Centripetal acceleration, a_c, is the acceleration directed towards the centre of the circle.

C. Centripetal force, F_c

Centripetal force, F_c, is the resultant force towards the centre that produces centripetal acceleration.

It is not a new kind of force: it is provided by real forces like tension, friction, normal reaction components, or gravity.

Velocity and acceleration directions in uniform circular motionA particle at the right side of a circular path has velocity tangent to the path and centripetal acceleration and resultant force directed horizontally inward toward the centre.centrervelocity vtangent to patha and ΣFinwardv ⟂ adirection changesspeed stays constant
Scroll diagram horizontally to read all labels.
At every point, velocity is tangent to the path while centripetal acceleration and the resultant force point towards the centre.

2. Key Ideas (What Earns Marks)

  • Even if speed is constant, velocity changes direction, so there is acceleration.
  • Centripetal acceleration: a_c = v²/r = rω²
  • Centripetal force (resultant inward force): F_c = ma_c = mv²/r = mrω²
  • In circular-motion force questions, write a radial equation: ∑ F_(towards centre) = mv²/r
  • Centripetal force acts perpendicular to velocity, so it does no work on the object (speed stays constant unless there is a tangential force).
Exam pitfall: treating centripetal force as an extra force

Centripetal force is the inward resultant of real forces already on the object. Do not add a separate Fc on top of tension, friction, normal reaction, or weight components.

3. Detailed Explanations

A. Why there is acceleration at constant speed

Velocity is a vector. In uniform circular motion, the direction of the velocity changes continuously, so velocity changes even if speed is constant.

Therefore the object accelerates.

B. Deriving a_c = v²/r (exam-safe sketch)

Over a small time Δ t, the velocity vector turns by a small angle Δθ.

The velocity magnitude stays v, but its direction changes, producing a change in velocity Δ v.

For small angles, Δ v ≈ vΔθ and Δθ/Δ t = ω.

So:

a = (Δ v)/(Δ t) ≈ vΔθ/(Δ t) = vω

Using v = rω gives:

a = vω = v(v/r) = v²/r

C. Centripetal force is the resultant radial force

Once you know a_c, Newton’s 2nd law gives:

∑ F_radial = ma_c

So the key skill is identifying which real forces have components towards the centre.

The radial equation describes the component of acceleration perpendicular to the instantaneous velocity. If speed also changes, analyse the tangential component separately; v²/r is still the instantaneous inward component for motion along a circular path.

Do not draw “centripetal force” as an extra arrow

Draw the real forces (tension, weight, normal, friction). Their resultant towards the centre is F_c.

D. Visual: why a_c grows quickly with speed

For a fixed radius, a_c = v²/r so a_c increases with the square of speed.

Centripetal acceleration vs speed (fixed radius)

A curved plot showing centripetal acceleration increasing with the square of speed.

Scroll across the graph to read all labels.

A curved plot showing centripetal acceleration increasing with the square of speed.A curved plot showing centripetal acceleration increasing with the square of speed.
Example with r = 2.0 m: a_c = v²/r. Doubling v makes a_c four times bigger. To linearise experimental data, plot a_c against v².
Open full-size graph
View figure data
Values for Centripetal acceleration vs speed (fixed radius)
Speed, v (m s⁻¹)Example (r = 2.0 m)
00
22
48
618
832

4. Common Mistakes

  • Drawing “centripetal force” as an extra force on the free-body diagram.
  • Using ∑ F = 0 because speed is constant (constant speed does not mean zero resultant force).
  • Putting forces in the wrong direction (centripetal acceleration is always towards the centre).
  • Mixing up r (radius) with diameter.

5. Exam Tips

  • Choose “towards the centre” as positive and write ∑ F_radial = mv²/r.
  • State which force (or component) provides the centripetal force.
  • If the path is vertical, the direction “towards the centre” flips between top/bottom — redraw the radial direction each time.

6. Worked Examples

Modelled example 1

Find centripetal acceleration

Core

Problem

A toy car moves at 2.0 m s⁻¹ around a circle of radius 0.80 m. Find the centripetal acceleration and state its direction.
Study the worked solution

  1. Choose the radial relation

    Method

    Use a_c = v²/r.

    Reason

    Speed and radius are the supplied circular-motion variables.

    Working

    a_c = (2.0)²/0.80

  2. Calculate and direct

    Method

    Obtain 5.0 m s⁻² towards the centre.

    Reason

    Centripetal acceleration is radial and inward.

    Working

    a_c = 5.0 m s⁻² inward.

Guided practice 2

Find centripetal force

About 5 min

Problem

A 0.50 kg mass moves at 3.0 m s⁻¹ in a horizontal circle of radius 1.2 m. Find the inward resultant force.

Try this before viewing the solution

Unit: N

Hints

Hint 1: force follows acceleration
Write ∑ F_radial = ma_c = mv²/r.
View solution step by step

  1. Build the radial equation

    Method

    Set the inward resultant to mv²/r.

    Reason

    Newton’s second law applies in the radial direction.

    Working

    F_radial = mv²/r

  2. Evaluate

    Method

    Substitute the data.

    Reason

    All quantities are in SI units.

    Working

    F_radial = 0.50(3.0)²/1.2 = 3.75 N

Common misconception 3

Tension provides centripetal force (horizontal circle)

Find and correct the mistake

Learner model

A 0.20 kg stone moves in a horizontal circle of radius 0.60 m at 5.0 m s⁻¹. A learner draws inward tension plus a separate inward “centripetal force”. Locate the first error and find the tension, assuming it supplies the radial resultant.

Try this before viewing the solution

First error

View solution step by step

  1. Correct the force model

    Method

    Keep tension as the real inward force and do not add another force.

    Reason

    “Centripetal force” is the radial resultant required by the motion.

    Working

    T = ∑ F_radial.

  2. Calculate tension

    Method

    Set T = mv²/r.

    Reason

    The prompt states tension alone supplies the radial resultant.

    Working

    T = 0.20(5.0)²/0.60 = 8.33 N

Examiner practice 4

Maximum speed on a flat track

4 marks

Examination question

A 1200 kg car takes a level bend of radius 60 m. Maximum horizontal friction is 7.2 kN. Find the maximum speed and identify the radial force. [4 marks]

Try this before viewing the solution

View solution step by step

  1. Identify radial force

    1 mark

    Method

    Use friction as the inward resultant at the limiting speed.

    Reason

    Weight and normal reaction balance vertically.

    Working

    F_friction = mv²/r.

  2. Convert force

    1 mark

    Method

    Write 7.2 kN = 7200 N.

    Reason

    SI force units are required.

    Working

    F = 7200 N.

  3. Rearrange

    1 mark

    Method

    Use v = square root of (Fr/m).

    Reason

    Speed appears squared in the radial equation.

    Working

    v = square root of (7200(60)/1200).

  4. Evaluate

    1 mark

    Method

    Obtain 19.0 m s⁻¹.

    Reason

    This is the speed at the friction limit.

    Working

    vₘₐₓ = 19.0 m s⁻¹.

Challenge 5

Vertical circle (tension at the top)

Minimal support

Changing-direction transfer

A 0.50 kg ball moves at 6.0 m s⁻¹ at the top of a vertical circle of radius 1.2 m. Find string tension; take g = 9.81 m s⁻².

Try this before viewing the solution

Hints

Hint 1: redraw inward at the top
At the top, both tension and weight point towards the centre.
View solution step by step

  1. Build the top radial equation

    Method

    Take downward/inward as positive and include both T and mg.

    Reason

    Both real forces point towards the centre at the top.

    Working

    T + mg = mv²/r

  2. Solve for tension

    Method

    Subtract weight from the required radial resultant.

    Reason

    Gravity supplies part of the inward force.

    Working

    T = 0.50(6.0)²/1.2-0.50(9.81) = 10.1 N

7. Mind Stretchers

Mind stretcher 1: Why doesn’t F_c change the speed?Extension

Explain why the centripetal force does no work in uniform circular motion.

Show Answer

Work done depends on the component of force along the displacement.

In uniform circular motion, the centripetal force is radial while the displacement/velocity is tangential, so they are perpendicular. The component along the motion is zero, so no work is done and speed stays constant (unless another tangential force acts).

Mind stretcher 2: If the string breaks…Extension

A ball on a string is moving in a circle. The string snaps.

Describe the direction of the ball’s motion immediately after.

Show Answer

It moves off in a straight line tangent to the circle at the point the string snaps (in the direction of its instantaneous velocity).

Mind stretcher 3: Optional (Enrichment)Extension

A. Banking (component of normal reaction)

On a banked track, a component of the normal reaction can provide the centripetal force.

If the question gives a banking angle, you resolve forces and apply ∑ F_radial = mv²/r.

B. Vertical circles (tension varies)

In vertical circular motion, the speed and the required centripetal force can change with position.

Problems often combine:

  • radial force equation ∑ F_radial = mv²/r, and
  • energy conservation between positions.

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027